The given numbers $2^{40}$, $3^{21}$, $4^{18}$, and $8^{12}$ can be simplified as follows:

$2^{40}$ remains as $2^{40}$ $3^{21}$ remain as $3^{21}$ $4^{18} = 2^{36}$ $8^{12} = 2^{36}$

This simplifies our choices to $2^{40}$ , $3^{21}$, $2^{36}$ , and $2^{36}$. Since $2^{40}$ is obviously larger than $2^{36}$ , it cannot be the smallest. Additionally, because we cannot have two correct answers, we are left to consider $3^{21}$ as the smallest option.

If further comparison were needed between $2^{36}$ and $3^{21}$, we could factorize them for an estimate without full calculation:

$2^{36}$ can be written as $2 \times 2^{35} = 2 \times (2^{5})^{7} = 2 \times 32^{7}$ $3^{21} = (3^{3})^{7} = 27^{7}$

From observation, $3^{21}$ is smaller than $2^{36}$ without detailed computation, as the exponential growth in the latter is more significant.